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Here is a very straightforward contre-example. Let $X=\mathbb CP^2\times \mathbb CP^1$ blown up in one point. Denote by $E$ the exceptional divisor, and denote by $\pi$ the projection of $X$ to $\mathbb CP^2$, and take the following bundle:

$$L_n=\pi^*(O(n))\otimes O(E),$$ where $n$ satisfies two conditions: $$c_1(O(n))^2\cdot c_1(A)>c_1(O(E))^2\cdot c_1(A),\;\;\;\; c_1(O(n))\cdot c_1(A)^2>c_1(O(E))\cdot c_1(A)^2,$$ it is obvious that such $n$ exists.

To that this bundle is what you want we just need the following two simple facts: $c_1(\pi^*(O(n)))\cdot c_1(O(E))=0$ and $c_1(O(E))^3=1$. $L_{n}$ is not big because the $H^0(kL_n)$ grows quadratically with $k$.

Idea of this example works in dimensions $2m+1$. We take a semi-ample line bundle $L_{sa}$ with Itaka dimension $2m$ http://en.wikipedia.org/wiki/Iitaka_dimension (in particular it is not big), and tensor it with a line bundle corresponding to an exceptional divisor $E$. We chose them so that $c_1(L_{sa})\cdot c_1(O(E))=0$, i.e., these bundles "don't interact". For large $n$ the class $(c_1(nL_{sa}))^k$ (provided $k\le 2m$) is represented by a cycle of a high degree (with respect to $A$), so it "beats" $(c_1(A))^k$. Finally $E^{2m+1}=1$.

Here is a very straightforward contre-example. Let $X=\mathbb CP^2\times \mathbb CP^1$ blown up in one point. Denote by $E$ the exceptional divisor, and denote by $\pi$ the projection of $X$ to $\mathbb CP^2$, and take the following bundle:

$$L_n=\pi^*(O(n))\otimes O(E),$$ where $n$ satisfies two conditions: $$c_1(O(n))^2\cdot c_1(A)>-c_1(O(E))^2\cdot c_1(A),\;\;\;\; c_1(O(n))\cdot c_1(A)^2>-c_1(O(E))\cdot c_1(A)^2,$$ it is obvious that such $n$ exists.

To that this bundle is what you want we just need the following two simple facts: $c_1(\pi^*(O(n)))\cdot c_1(O(E))=0$ and $c_1(O(E))^3=1$. $L_{n}$ is not big because the $H^0(kL_n)$ grows quadratically with $k$.

Idea of this example works in dimensions $2m+1$. We take a semi-ample line bundle $L_{sa}$ with Itaka dimension $2m$ http://en.wikipedia.org/wiki/Iitaka_dimension (in particular it is not big), and tensor it with a line bundle corresponding to an exceptional divisor $E$. We chose them so that $c_1(L_{sa})\cdot c_1(O(E))=0$, i.e., these bundles "don't interact". For large $n$ the class $(c_1(nL_{sa}))^k$ (provided $k\le 2m$) is represented by a cycle of a high degree (with respect to $A$), so it "beats" $(c_1(O(E)))^k$. Finally $E^{2m+1}=1$.

Here is an example. Let $X=\mathbb CP^2\times \mathbb CP^1$ blown up in one point. Denote by $E$ the exceptional divisor, and denote by $\pi$ the projection of $X$ to $\mathbb CP^2$. Then for $n=n(A)$ sufficiently large $L_n=\pi^*(O(n))\otimes O(E)$ is what you want. To see this we need the following two simple facts: $c_1(\pi^*(O(n)))\cdot c_1(O(E))=0$ and $c_1(O(E))^3=1$. $L_{n}$ is not big because the $H^0(kL_n)$ grows quadratically with $k$.

The idea of this example is very simple, it works in dimensions $2m+1$. We take a semi-ample line bundle $L_{sa}$ with Itaka dimension $2m$ http://en.wikipedia.org/wiki/Iitaka_dimension (in particular it is not big), and tensor it with a line bundle corresponding to an exceptional divisor $E$. We chose them so that $c_1(L_{sa})\cdot c_1(O(E))=0$, i.e., these bundles "don't interact". For large $n$ the class $(c_1(nL_{sa}))^k$ (provided $k\le 2m$) is represented by a cycle of a high degree (with respect to $A$), so it "beats" $(c_1(A))^k$. Finally $E^{2m+1}=1$.

Here is a very straightforward contre-example. Let $X=\mathbb CP^2\times \mathbb CP^1$ blown up in one point. Denote by $E$ the exceptional divisor, and denote by $\pi$ the projection of $X$ to $\mathbb CP^2$, and take the following bundle:

$$L_n=\pi^*(O(n))\otimes O(E),$$ where $n$ satisfies two conditions: $$c_1(O(n))^2\cdot c_1(A)>c_1(O(E))^2\cdot c_1(A),\;\;\;\; c_1(O(n))\cdot c_1(A)^2>c_1(O(E))\cdot c_1(A)^2,$$ it is obvious that such $n$ exists.

To that this bundle is what you want we just need the following two simple facts: $c_1(\pi^*(O(n)))\cdot c_1(O(E))=0$ and $c_1(O(E))^3=1$. $L_{n}$ is not big because the $H^0(kL_n)$ grows quadratically with $k$.

Idea of this example works in dimensions $2m+1$. We take a semi-ample line bundle $L_{sa}$ with Itaka dimension $2m$ http://en.wikipedia.org/wiki/Iitaka_dimension (in particular it is not big), and tensor it with a line bundle corresponding to an exceptional divisor $E$. We chose them so that $c_1(L_{sa})\cdot c_1(O(E))=0$, i.e., these bundles "don't interact". For large $n$ the class $(c_1(nL_{sa}))^k$ (provided $k\le 2m$) is represented by a cycle of a high degree (with respect to $A$), so it "beats" $(c_1(A))^k$. Finally $E^{2m+1}=1$.

Here is an example. Let $X=\mathbb CP^2\times \mathbb CP^1$ blown up in one point. Denote by $E$ the exceptional divisor, and denote by $\pi$ the projection of $X$ to $\mathbb CP^2$. Then for $n=n(A)$ sufficiently large $L_n=\pi^*(O(n))\otimes O(E)$ is what you want. To see this we need the following two simple facts: $c_1(\pi^*(O(n)))\cdot c_1(O(E))=0$ and $c_1(O(E))^3=1$. $L_{n}$ is not big because the $H^0(kL_n)$ grows quadratically with $k$.

Here is an example. Let $X=\mathbb CP^2\times \mathbb CP^1$ blown up in one point. Denote by $E$ the exceptional divisor, and denote by $\pi$ the projection of $X$ to $\mathbb CP^2$. Then for $n=n(A)$ sufficiently large $L_n=\pi^*(O(n))\otimes O(E)$ is what you want. To see this we need the following two simple facts: $c_1(\pi^*(O(n)))\cdot c_1(O(E))=0$ and $c_1(O(E))^3=1$. $L_{n}$ is not big because the $H^0(kL_n)$ grows quadratically with $k$.

The idea of this example is very simple, it works in dimensions $2m+1$. We take a semi-ample line bundle $L_{sa}$ with Itaka dimension $2m$ http://en.wikipedia.org/wiki/Iitaka_dimension (in particular it is not big), and tensor it with a line bundle corresponding to an exceptional divisor $E$. We chose them so that $c_1(L_{sa})\cdot c_1(O(E))=0$, i.e., these bundles "don't interact". For large $n$ the class $(c_1(nL_{sa}))^k$ (provided $k\le 2m$) is represented by a cycle of a high degree (with respect to $A$), so it "beats" $(c_1(A))^k$. Finally $E^{2m+1}=1$.